Let us first show that \(\mathcal {C}\)  is a \([2^t+\lceil \frac{2^t}{r}\rceil , 2^t-1-t,5;r]\) -locally repairable code. By Proposition REF , it is sufficient to show that any four columns of \(H\)  are linearly independent. Let us prove this only for the case where \(r=3\) .
Case 1: all four columns belong to the same block, say the first block. In this case, we have to show that \((1,{\bf 0},{\bf 0})^T\) , \((1,{\bf 0},{\bf h}_{11})^T\) , \((1,{\bf 0},{\bf h}_{12})^T\)  and \((1,{\bf 0},{\bf h}_{13})^T\)  are linearly independent. Suppose that
\(\lambda _0(1,{\bf 0},{\bf 0})^T+\sum _{i=1}^3\lambda _i(1,{\bf 0},{\bf h}_{1i})^T={\bf 0}^T\) 
for some \(\lambda _i\in \mathbb {F}_2\)  with \(0\leqslant i\leqslant 3\) . Then we have \(\sum _{i=1}^3\lambda _i{\bf h}_{1i}={\bf 0}\) . This forces that \(\lambda _1=\lambda _2=\lambda _3=0\)  from [1]}, since \(\mathcal {C}_0\)  has minimum distance 4. Thus, it follows from (REF ) that \(\lambda _0=0\) .
Case 2: Other possibilities: (a) three columns belong to one block and one column belongs to another block; (b) two columns belong to one block and other two belong to another block; (c) two columns belong to one block and other two columns belong to other two blocks, respectively; (d) four columns belong to four distinct blocks. For these four possibilities, one can use a similar argument to show that they are linearly independent. This proves the first part.
Now if \(r=2\)  and \(t\geqslant 5\) , we want to show that
\(k=2^t-1-t\geqslant \left\lfloor \frac{rn}{r+1}-\min \left\lbrace \log _2\left(1+\frac{rn}{2}\right),\frac{rn}{(r+1)(r+2)}\right\rbrace \right\rfloor \) 
with \(r=2\)  and \(n=2^t+2^{t-1}\) .
To show (REF ), it will be sufficient to show the following two inequalities
\(2^t-t>\frac{rn}{r+1}-\log _2\left(1+\frac{rn}{2}\right)\) 
and
\(2^t-t>\frac{rn}{r+1}-\frac{rn}{(r+1)(r+2)}.\) 
Substituting \(r=2\)  and \(n=2^t+2^{t-1}\)  into (REF ) and rewrite it into the following form
\(\log _2\left(1+2^t+2^{t-1}\right)>t.\) 
It is clear that (REF ) holds and hence (REF ) holds.
Substituting \(r=2\)  and \(n=2^t+2^{t-1}\)  into (REF ) and rewrite it into the following form
\(2^t-t>\frac{r}{r+2}\times n=2^{t-1}+2^{t-2}.\) 
The inequality (REF ) holds when \(t\geqslant 5\)  and hence (REF ) holds. This completes the proof for the case where \(r=2\)  and \(t\geqslant 5\) .
Next assume that \(r=3\)  and \(t\geqslant 5\) . In this case we want to show that the two inequalities (REF ) and (REF )
hold for \(r=3\)  and \(n=2^t+\lceil \frac{2^{t}}{3}\rceil \) .
Substituting \(r=3\)  and \(n=2^t+\left\lceil \frac{2^{t}}{3}\right\rceil \)  into (REF ) and rewrite it into the following form
\(\log _2\left(1+\frac{3}{2}\left(2^t+\left\lceil \frac{2^{t}}{3}\right\rceil \right)\right)>t+\frac{3}{4}\times \left\lceil \frac{2^{t}}{3}\right\rceil -\frac{2^{t}}{4}.\) 
Note that we have \(1+\frac{3}{2}\left(2^t+\left\lceil \frac{2^{t}}{3}\right\rceil \right)>\frac{3}{2}\times 2^t>\sqrt{2}\times 2^t\) . This gives
\(\log _2\left(1+\frac{3}{2}\left(2^t+\left\lceil \frac{2^{t}}{3}\right\rceil \right)\right)>t+\frac{1}{2}=t+\frac{3}{4}\times \frac{2^{t}+2}{3}-\frac{2^{t}}{4}\geqslant t+\frac{3}{4}\times \left\lceil \frac{2^{t}}{3}\right\rceil -\frac{2^{t}}{4},\) 
i.e., the inequality (REF ) holds and hence (REF ) holds.
Substituting \(r=3\)  and \(n=2^t+\left\lceil \frac{2^{t}}{3}\right\rceil \)  into (REF ) and rewrite it into the following form
\(2^t-t>\frac{r}{r+2}\times n=\frac{3}{5}\left(2^t+\left\lceil \frac{2^{t}}{3}\right\rceil \right)\) 
The inequality (REF ) holds when \(t\geqslant 5\)  and hence (REF ) holds. This completes the proof for the case where \(r=3\)  and \(t\geqslant 5\) .
